Problem: $f(x) = \begin{cases} 5 & \text{if } x = 0 \\ 2x^{2}-1 & \text{otherwise} \end{cases}$ What is the range of $f(x)$ ?
Solution: First consider the behavior for $x \ne 0$ Consider the range of $2x^{2}$ The range of $x^2$ is $\{\, y \mid y \ge 0 \,\}$ Multiplying by $2$ doesn't change the range. To get $2x^{2}-1$ , we subtract $1$ So the range becomes: $\{\, y \mid y ≥ -1 \,\}$ If $x = 0$ , then $f(x) = 5$ , which eliminates $f(x) = -1$ from the range. The new range is $\{\, y \mid y > -1 \,\}$.